Bài 1: Tìm ĐKXĐ
1, \(\sqrt{2x-1}\)+ \(\frac{\sqrt{x-3}}{\sqrt{5-x}}\)
2, \(\sqrt{x-1}\)+ \(\frac{\sqrt{2-x}}{\sqrt{x+1}}\)
Bài 2: Tính
1, A = \(\sqrt{6+2\sqrt{5}}\) - \(\sqrt{6-2\sqrt{5}}\)
2, B = \(\sqrt{4+\sqrt{15}}\) + \(\sqrt{4-\sqrt{15}}\) - \(2\sqrt{3-\sqrt{5}}\)
3, C = \(\sqrt{4+\sqrt{10}+2\sqrt{5}}\) + \(\sqrt{4-\sqrt{10}+2\sqrt{5}}\)
4, D = \(\sqrt{15-6\sqrt{6}}\) + \(\sqrt{15+6\sqrt{6}}\)
5, E = \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
6, F = \(\sqrt{\left(1-\sqrt{2021}\right)}\). \(\sqrt{2022+2\sqrt{2021}}\)
7, G = \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\) + \(\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
Bài 1:
1. ĐKXĐ: \(\left\{\begin{matrix} 2x-1\geq 0\\ x-3\geq 0\\ 5-x>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ x\geq 3\\ x< 5\end{matrix}\right.\Leftrightarrow 3\leq x< 5\)
2.
ĐKXĐ: \(\left\{\begin{matrix} x-1\geq 0\\ 2-x\geq 0\\ x+1>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\leq 2\\ x>-1\end{matrix}\right.\Leftrightarrow 1\leq x\leq 2\)
2.1
\(A=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5.1}+1}-\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}=|\sqrt{5}+1|-|\sqrt{5}-1|=2\)
2.2
\(B\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{3+2\sqrt{3.5}+5}+\sqrt{3-2\sqrt{3.5}+5}-2\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5})^2}+\sqrt{(\sqrt{3}-\sqrt{5})^2}-2\sqrt{(\sqrt{5}-1)^2}\)
\(=|\sqrt{3}+\sqrt{5}|+|\sqrt{3}-\sqrt{5}|-2|\sqrt{5}-1|=2\)
$\Rightarrow B=\sqrt{2}$
2.3
$C^2=8+4\sqrt{5}+2\sqrt{(4+\sqrt{10}+2\sqrt{5})(4-\sqrt{10}+2\sqrt{5})}$
$=8+4\sqrt{5}+2\sqrt{(4+2\sqrt{5})^2-(\sqrt{10})^2}$
$=8+4\sqrt{5}+2\sqrt{26+16\sqrt{5}}$
$\Rightarrow C=\sqrt{8+4\sqrt{5}+2\sqrt{26+16\sqrt{5}}}$
(mình nghĩ bạn viết nhầm đề?)
2.4
$D=\sqrt{15-2\sqrt{54}}+\sqrt{15+2\sqrt{54}}$
$=\sqrt{6-2\sqrt{6.9}+9}+\sqrt{6+2\sqrt{6.9}+9}$
$=\sqrt{(\sqrt{6}-3)^2}+\sqrt{(\sqrt{6}+3)^2}$
$=|\sqrt{6}-3|+|\sqrt{6}+3|=6$
2.5
\(E=\frac{(\sqrt{2}+\sqrt{3}+2)+(2+\sqrt{6}+\sqrt{8})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+(\sqrt{4}+\sqrt{6}+\sqrt{8})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
2.6
\(F=\sqrt{(1-\sqrt{2021})^2}\sqrt{2021+2\sqrt{2021.1}+1}\)
\(=|1-\sqrt{2021}|\sqrt{\sqrt{2021}+1)^2}=|1-\sqrt{2021}||1+\sqrt{2021}|\)
\(=|1-2021|=2020\)
2.7
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